# Elimination of OH from spirocyclic compounds

I was wondering, what would happen if I treated the following compound i.e. trispiro[2.1.25.1.29.13]dodecane-4,8,12-triol with sulfuric acid[$$\ce{H_2SO_4}$$] to eliminate OH group to form double bond:

Will the double bond form? if yes then where and why? If no then what other compound will form and why?

• Well, there's gonna be various rearrangements Oct 8, 2016 at 12:42
• Agree, probably a lot of 1,2-alkyl shifts. Oct 8, 2016 at 13:04
• You might get a carbonyl from a hydride shift...
– Zhe
Oct 8, 2016 at 15:59
• Have you tried to draw the structure of a compound resulting from elimination? Oct 19, 2016 at 15:14
• @jerepierre I am really confused about what would happen. I think that there will be a ring expansion for each of those cyclopropane rings, forming a four member bridged ring. But then where would the double bonds form? Bredt's rule disallows formation of double bonds at bridgehead. Oct 20, 2016 at 8:47

This question has been unanswered for quite long, and meanwhile, I have found an answer, so I wanted to share it.

On treating the compound with sulfuric acid, the $\ce{OH}$ groups will be protonated to $\ce{OH_2^+}$.

The $\ce{OH_2^+}$ groups are good leaving groups, so they will leave as $\ce{H_2O}$, forming carbocation.(one by one)

As cyclopropane is under significant ring strain, it will undergo ring expansion to form cyclobutane ring, which is somewhat less strained. At the same time, $\ce{H^+}$ ions will be expelled, forming double bonds. (It should also take place step by step.)

In this case I don't think that the double bond will not cause any significant barrier to the formation of the molecules, as the double bond is here the bridging bond, not at the bridgehead[formation of double bonds at bridgehead is not possible for smaller rings(<8 carbon rings), as it causes very high strain: Bredt's rule]

I have considered only elimination in this reaction scheme. Other compounds may also be formed. Feel free to post answers if you think other compounds will be formed or if you think the above mechanism is wrong in any way.

• I don’t think the steps would happen to each carbon simultaneously but other than that this sounds reasonable.
– Jan
Dec 3, 2016 at 18:46
• @Jan Why not simultaneously? Dec 4, 2016 at 4:32
• Carbotrications seem like a far-fetched idea to me... Dec 4, 2016 at 7:10
• sequential not simultaneous reaction Dec 5, 2016 at 15:24
• Double bonds won't form in this scenario. It still would be a triol probably. Dec 5, 2016 at 15:50